Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $x = \dfrac{-5a + 15}{a^2 + 3a - 54} \times \dfrac{5a - 30}{2a - 6} $
First factor the quadratic. $x = \dfrac{-5a + 15}{(a - 6)(a + 9)} \times \dfrac{5a - 30}{2a - 6} $ Then factor out any other terms. $x = \dfrac{-5(a - 3)}{(a - 6)(a + 9)} \times \dfrac{5(a - 6)}{2(a - 3)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ -5(a - 3) \times 5(a - 6) } { (a - 6)(a + 9) \times 2(a - 3) } $ $x = \dfrac{ -25(a - 3)(a - 6)}{ 2(a - 6)(a + 9)(a - 3)} $ Notice that $(a - 3)$ and $(a - 6)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ -25(a - 3)\cancel{(a - 6)}}{ 2\cancel{(a - 6)}(a + 9)(a - 3)} $ We are dividing by $a - 6$ , so $a - 6 \neq 0$ Therefore, $a \neq 6$ $x = \dfrac{ -25\cancel{(a - 3)}\cancel{(a - 6)}}{ 2\cancel{(a - 6)}(a + 9)\cancel{(a - 3)}} $ We are dividing by $a - 3$ , so $a - 3 \neq 0$ Therefore, $a \neq 3$ $x = \dfrac{-25}{2(a + 9)} ; \space a \neq 6 ; \space a \neq 3 $